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Program to find leftmost deepest node of a tree in Python
Suppose we have a binary tree; we have to find the value of the deepest node. If there are more than one deepest node, then return the leftmost deepest node.
So, if the input is like
then the output will be 4 as 4 and 7 are deepest but 4 is left most.
To solve this, we will follow these steps −
queue := a queue with one node root.
left_max := value of root
while size of queue > 0, do
level_size := size of queue
for i in range 0 to level_size, do
temp := first element from queue and delete it
if i is same as 0, then
left_max := value of temp
if left of temp is non-zero, then
insert left of temp at the end of queue
if right of temp is not null, then
insert right of temp at the end of queue
return left_max
Let us see the following implementation to get better understanding −
Example
class TreeNode: def __init__(self, value): self.val = value self.left = None self.right = None class Solution: def solve(self, root): queue = [root] left_max = root.val while len(queue) > 0: level_size = len(queue) for i in range(level_size): temp = queue.pop(0) if i == 0: left_max = temp.val if temp.left: queue.append(temp.left) if temp.right: queue.append(temp.right) return left_max ob = Solution() root = TreeNode(13) root.left = TreeNode(12) root.right = TreeNode(14) root.right.left = TreeNode(16) root.right.right = TreeNode(22) root.right.left.left = TreeNode(4) root.right.left.right = TreeNode(7) print(ob.solve(root))
Input
root = TreeNode(13) root.left = TreeNode(12) root.right = TreeNode(14) root.right.left = TreeNode(16) root.right.right = TreeNode(22) root.right.left.left = TreeNode(4) root.right.left.right = TreeNode(7)
Output
4
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